Solubility Product Constant A special case of equilibrium involving dissolving. Solid  Positive Ion + Negative Ion Mg(NO 3 ) 2  Mg NO 3 - Keq = [Mg 2+ ] [NO 3 - ] 2 Because the constant is a product of solubility, we call it the solubility product constant Ksp

Solubility Product Problems Given Ksp, find solubility Given solubility, find Ksp Find solubility in a solution with a common ion Predicting precipitation

Solubility Product Constant (Ksp) BaF 2(S)  Ba +2 (aq) + 2F - (aq) Write out the equilibrium law expression… Ksp = [Ba +2 ][F - ] 2

Solubility Generalizations All nitrates are soluble All compounds of the alkali metals are soluble (Li, Na, K, etc.) All compounds of the ammonium (NH 4 + ) are soluble

Given Ksp, Find Solubility What is the solubility of Silver Bromide (Ksp = 5.2 x ) AgBr  Ag + + Br - Ksp = [Ag + ][Br - ] = 5.2 x Let x = the solubility (x)(x) = 5.2 x X 2 = 5.2 x X = 7.2 x 10 -7

Another Example What is the solubility of PbI 2 (Ksp = 7.1 x ) PbI 2(s)  Pb I - Ksp = [Pb +2 ][I - ] 2 = 7.1 x Let x = the solubility (x)(2x) 2 = 7.1 x (x)(4x 2 ) = 7.1 x x 3 = 7.1 x X = 1.2x10 -3 M

Find Ksp Given Solubility What is the Ksp of Boric Acid, given its solubility of 2.15 x Moles/liter? H 3 BO 3  3H + + BO 3 -3 Ksp = [H + ] 3 [BO 3 -3 ] [3(2.15x10 -3 )] 3 [2.15x10 -3 ] = 5.77 x

Solubility with a Common Ion What is the solubility of lead iodide (PbI 2 ) in a.15M solution of KI ? PbI 2  Pb I - KI  K + + I - Ksp = [Pb +2 ][I - ] 2 = 7.1 x 10 -9

Solubility with a Common Ion What is the solubility of lead iodide (PbI 2 ) in a.15M solution of KI ? PbI 2  Pb I - KI  K + + I - Ksp = [Pb +2 ][I - ] 2 Let x = solubility Then: [Pb+] = x [I-] =.15+2x

Ksp = [Pb +2 ][I - ] 2 = 7.1 x Let x = solubility of PbI 2 in the solution Then, [Pb +2 ] = x [I - ] = x Ksp = (x)(.15+2x) 2 = 7.1 x 10 -9

Ksp = [Pb +2 ][I - ] 2 = Ksp = (x)(.15+2x) 2 = (x)(4x 2 +.6x ) = x 3 +.6x x = x 3 +.6x – = 0

Ksp = [Pb +2 ][I - ] 2 = 7.1 x Ksp = (x)(.15+2x) 2 = 7.1 x Assume.15>>2x then,.15+2x .15

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Ksp = [Pb +2 ][I - ] 2 = 7.1 x Ksp = (x)(.15+2x) 2 = 7.1 x Assume.15>>2x then,.15+2x .15 Ksp = (x)(.15) 2 = 7.1 x X = 3.16 x moles/liters

Predicting Precipitation A student mixes mole Ca(NO 3 ) 2 in 2 liters of 0.10M Na 2 CO 3 solution. Will a precipitate form? Step 1: Write out the dissolving equations Ca(NO 3 ) 2 (s)  Ca 2+ (aq) + 2NO 3 - (aq) Na 2 CO 3 (s)  2Na + (aq) + CO 3 2- (aq)

Step 2: Determine the most likely precipitate & write out it’s equation. Ca(NO 3 ) 2  Ca NO 3 - Na 2 CO 3  2Na + + CO 3 2-

Recall the solubility generalizations… All nitrates are soluble All compounds of the alkali metals are soluble (Li, Na, K, etc.) All compounds of the ammonium (NH 4 + ) are soluble

Step 2: Determine the most likely precipitate & write out it’s equation. Ca(NO 3 ) 2  Ca NO 3 - Na 2 CO 3  2Na + + CO 3 2- CaCO 3  Ca 2+ + CO 3 2-

Ksp = [Ca 2+ ][CO 3 2- ] = 4.7 x Step 3: Determine the molar concentrations & calculate the reaction quotient (Q). [Ca 2+ ] =.01 mole/2 liters =.005M [CO 3 2- ] = 0.10 M (given)

Reaction Quotient (Q) The product of the Ksp equation using the ion concentration before any reaction interaction. If Q > Ksp Then a precipitate will form.

Q = [Ca 2+ ][CO 3 2- ] [Ca 2+ ] =.01mole/2 liters =.005M [CO 3 2- ] = 0.10 M (given) Q = (.005)(0.10) =.0005 Q > Ksp  a precipitate will form.

You try one…..015 moles of AgNO 3 is mixed with 5 liters of.02M NaCl solution. What is the most likely precipitate and will it form?

You try one…..015 moles of AgNO 3 is mixed with 5 liters of.02M NaCl solution. What is the most likely precipitate and will it form? AgNO 3  Ag + + NO 3 - NaCl  Na + + Cl - AgCl  Ag + + Cl -

Q = [Ag + ][Cl - ] Q = (.015/5)(.02) = Look up the Ksp (1.0 x ) Q > Ksp So a precipitate will form!